Netzahualcoyotl Guadarrama Camarena
Netzahualcoyotl Guadarrama Camarena Información personal Nombre
Netzahualcoyotl Guadarrama Camarena Nacimiento
26 de Octubre de 1977Tierra . Sexo
Masculino Ocupación
Wikipedista Sitio web
Hola a todos. Soy un usuario que colaboro en este proyecto.
Yo soy colaborador en wikipedia, nací en Mexico. Soy colaborador(a) de Wikipedia desde 2005.
Mi personalidad [ editar ] Calculo Vectorial
Gradiente en coordenadas generalizadas usando k-forma diferenciales [ editar ]
d
z
=
∂
z
∂
(
q
1
,
q
2
,
.
.
.
,
q
n
)
d
(
q
1
,
q
2
,
.
.
.
,
q
n
)
=
[
H
1
H
^
1
H
2
H
^
2
.
.
.
H
n
H
^
n
]
d
q
{\displaystyle dz={\partial z \over \partial (q_{1},q_{2},...,q_{n})}d(q_{1},q_{2},...,q_{n})=[H_{1}{\widehat {H}}_{1}H_{2}{\widehat {H}}_{2}...H_{n}{\widehat {H}}_{n}]dq}
d
z
=
[
∂
z
H
1
∂
q
1
∂
z
H
2
∂
q
2
.
.
.
∂
z
H
n
∂
q
n
]
[
H
1
d
q
1
H
2
d
q
2
.
.
.
H
n
d
q
]
T
=
[
H
1
^
H
2
^
.
.
.
H
n
^
]
[
d
l
1
d
l
2
.
.
.
d
l
n
]
T
=
∇
l
z
d
l
{\displaystyle dz=[{\partial z \over H_{1}\partial q_{1}}{\partial z \over H_{2}\partial q_{2}}{...}{\partial z \over H_{n}\partial q_{n}}][{H_{1}dq_{1}}{H_{2}dq_{2}}{}...{H_{n}dq}]^{T}=[{\widehat {H_{1}}}{\widehat {H_{2}}}{...}{\widehat {H_{n}}}][{dl_{1}}{dl_{2}}{...}{dl_{n}}]^{T}=\nabla _{l}zdl}
Divergencia en coordenadas generalizadas usando k-forma diferenciales [ editar ]
η
=
ϕ
q
3
h
1
d
q
1
h
2
d
q
2
+
ϕ
q
1
h
2
d
q
2
h
3
d
q
3
+
ϕ
q
2
h
3
d
q
3
h
1
d
q
1
{\displaystyle \eta =\phi _{q_{3}}h_{1}dq_{1}h_{2}dq_{2}+\phi _{q_{1}}h_{2}dq_{2}h_{3}dq_{3}+\phi _{q_{2}}h_{3}dq_{3}h_{1}dq_{1}}
d
η
=
d
(
ϕ
q
3
h
1
h
2
∧
d
q
1
d
q
2
)
+
d
(
ϕ
q
1
h
2
h
3
∧
d
q
1
d
q
3
)
+
d
(
ϕ
q
1
h
3
h
1
∧
d
q
3
d
q
1
)
{\displaystyle d{\eta }=d(\phi _{q_{3}}h_{1}h_{2}\land dq_{1}dq_{2})+d(\phi _{q_{1}}h_{2}h_{3}\land dq_{1}dq_{3})+d(\phi _{q_{1}}h_{3}h_{1}\land dq_{3}dq_{1})}
d
η
=
d
(
ϕ
q
3
h
1
h
2
)
∧
d
q
1
d
q
2
+
(
ϕ
q
3
h
1
h
2
)
∧
d
(
d
q
1
d
q
2
)
+
d
(
ϕ
q
1
h
2
h
3
)
∧
d
q
2
d
q
3
+
(
ϕ
q
1
h
2
h
3
)
∧
d
(
d
q
2
d
q
3
)
+
d
(
ϕ
q
2
h
3
h
1
)
∧
d
q
3
d
q
1
+
(
ϕ
q
2
h
3
h
1
)
∧
d
(
d
q
3
d
q
1
)
{\displaystyle d{\eta }=d(\phi _{q_{3}}h_{1}h_{2})\land dq_{1}dq_{2}+(\phi _{q_{3}}h_{1}h_{2})\land d(dq_{1}dq_{2})+d(\phi _{q_{1}}h_{2}h_{3})\land dq_{2}dq_{3}+(\phi _{q_{1}}h_{2}h_{3})\land d(dq_{2}dq_{3})+d(\phi _{q_{2}}h_{3}h_{1})\land dq_{3}dq_{1}+(\phi _{q_{2}}h_{3}h_{1})\land d(dq_{3}dq_{1})}
d
η
=
(
∂
ϕ
q
3
h
1
h
2
∂
q
1
d
q
1
+
∂
ϕ
q
3
h
1
h
2
∂
q
2
d
q
2
+
∂
ϕ
q
3
h
1
h
2
∂
q
3
d
q
3
)
∧
d
q
1
d
q
2
+
(
∂
ϕ
q
1
h
1
h
2
∂
q
1
d
q
1
+
∂
ϕ
q
1
h
1
h
2
∂
q
2
d
q
2
+
∂
ϕ
q
1
h
1
h
2
∂
q
3
d
q
3
)
∧
d
q
2
d
q
3
+
(
∂
ϕ
q
2
h
1
h
2
∂
q
1
d
q
1
+
∂
ϕ
q
2
h
1
h
2
∂
q
2
d
q
2
+
∂
ϕ
q
2
h
1
h
2
∂
q
3
d
q
3
)
∧
d
q
3
d
q
1
{\displaystyle d{\eta }=({{\partial \phi _{q_{3}}h_{1}h_{2}} \over {\partial q_{1}}}dq_{1}+{{\partial \phi _{q_{3}}h_{1}h_{2}} \over {\partial q_{2}}}dq_{2}+{{\partial \phi _{q_{3}}h_{1}h_{2}} \over {\partial q_{3}}}dq_{3})\land dq_{1}dq_{2}+({{\partial \phi _{q_{1}}h_{1}h_{2}} \over {\partial q_{1}}}dq_{1}+{{\partial \phi _{q_{1}}h_{1}h_{2}} \over {\partial q_{2}}}dq_{2}+{{\partial \phi _{q_{1}}h_{1}h_{2}} \over {\partial q_{3}}}dq_{3})\land dq_{2}dq_{3}+({{\partial \phi _{q_{2}}h_{1}h_{2}} \over {\partial q_{1}}}dq_{1}+{{\partial \phi _{q_{2}}h_{1}h_{2}} \over {\partial q_{2}}}dq_{2}+{{\partial \phi _{q_{2}}h_{1}h_{2}} \over {\partial q_{3}}}dq_{3})\land dq_{3}dq_{1}}
d
η
=
(
∂
ϕ
q
1
h
2
h
3
∂
q
1
+
∂
ϕ
q
2
h
3
h
1
∂
q
2
+
∂
ϕ
q
3
h
1
h
2
∂
q
3
)
∧
d
q
1
d
q
2
d
q
3
{\displaystyle d{\eta }=({{\partial \phi _{q_{1}}h_{2}h_{3}} \over {\partial q_{1}}}+{{\partial \phi _{q_{2}}h_{3}h_{1}} \over {\partial q_{2}}}+{{\partial \phi _{q_{3}}h_{1}h_{2}} \over {\partial q_{3}}})\land dq_{1}dq_{2}dq_{3}}
d
η
=
1
h
1
h
2
h
3
(
∂
ϕ
q
1
h
2
h
3
∂
q
1
+
∂
ϕ
q
2
h
3
h
1
∂
q
2
+
∂
ϕ
q
3
h
1
h
2
∂
q
3
)
∧
h
1
d
q
1
h
2
d
q
2
h
3
d
q
3
{\displaystyle d{\eta }={1 \over h_{1}h_{2}h_{3}}({{\partial \phi _{q_{1}}h_{2}h_{3}} \over {\partial q_{1}}}+{{\partial \phi _{q_{2}}h_{3}h_{1}} \over {\partial q_{2}}}+{{\partial \phi _{q_{3}}h_{1}h_{2}} \over {\partial q_{3}}})\land h_{1}dq_{1}h_{2}dq_{2}h_{3}dq_{3}}
d
η
=
1
h
1
h
2
h
3
(
∂
ϕ
q
1
h
2
h
3
∂
q
1
+
∂
ϕ
q
2
h
3
h
1
∂
q
2
+
∂
ϕ
q
3
h
1
h
2
∂
q
3
)
∧
d
l
1
d
l
2
d
l
3
{\displaystyle d{\eta }={1 \over h_{1}h_{2}h_{3}}({{\partial \phi _{q_{1}}h_{2}h_{3}} \over {\partial q_{1}}}+{{\partial \phi _{q_{2}}h_{3}h_{1}} \over {\partial q_{2}}}+{{\partial \phi _{q_{3}}h_{1}h_{2}} \over {\partial q_{3}}})\land dl_{1}dl_{2}dl_{3}}
d
η
=
d
(
ϕ
→
⋅
d
s
→
)
=
∇
⋅
ϕ
→
d
v
{\displaystyle d{\eta }=d({\vec {\phi }}\centerdot d{\vec {s}})={\nabla \centerdot {\vec {\phi }}dv}}
Rotacional en coordenadas generalizadas usando k-forma diferenciales [ editar ]
ω
=
ϕ
l
1
d
l
1
+
ϕ
l
2
d
l
2
+
ϕ
l
3
d
l
3
=
ϕ
l
1
h
1
d
q
1
+
ϕ
l
2
h
2
d
q
2
+
ϕ
l
3
h
3
d
q
3
{\displaystyle {\omega }={\phi _{l_{1}}}dl_{1}+{\phi _{l_{2}}}dl_{2}+{\phi _{l_{3}}}dl_{3}={\phi _{l_{1}}}h_{1}dq_{1}+{\phi _{l_{2}}}h_{2}dq_{2}+{\phi _{l_{3}}}h_{3}dq_{3}}
d
ω
=
d
(
h
1
ϕ
l
1
∧
d
q
1
)
+
d
(
h
2
ϕ
l
2
∧
d
q
2
)
+
d
(
h
3
ϕ
l
3
∧
d
q
3
)
{\displaystyle {d\omega }=d(h_{1}\phi _{l_{1}}\land dq_{1})+d(h_{2}\phi _{l_{2}}\land dq_{2})+d(h_{3}\phi _{l_{3}}\land dq_{3})}
d
ω
=
d
h
1
ϕ
l
1
∧
d
q
1
+
h
1
ϕ
l
1
∧
d
d
q
1
+
d
h
2
ϕ
l
2
∧
d
q
2
+
h
2
ϕ
l
2
∧
d
d
q
2
+
d
h
3
ϕ
l
3
∧
d
q
1
+
h
3
ϕ
l
3
∧
d
d
q
3
{\displaystyle d\omega =d{h_{1}\phi _{l_{1}}}\land dq_{1}+{h_{1}\phi _{l_{1}}}\land ddq_{1}+d{h_{2}\phi _{l_{2}}}\land dq_{2}+{h_{2}\phi _{l_{2}}}\land ddq_{2}+d{h_{3}\phi _{l_{3}}}\land dq_{1}+{h_{3}\phi _{l_{3}}}\land ddq_{3}}
d
ω
=
[
∂
h
1
ϕ
l
1
∂
q
1
d
q
1
+
∂
h
1
ϕ
l
1
∂
q
2
d
q
2
+
∂
h
1
ϕ
l
1
∂
q
3
d
q
3
]
∧
d
q
1
+
[
∂
h
2
ϕ
l
2
∂
q
1
d
q
1
+
∂
h
2
ϕ
l
2
∂
q
2
d
q
2
+
∂
h
2
ϕ
l
2
∂
q
3
d
q
3
]
∧
d
q
2
+
[
∂
h
3
ϕ
l
3
∂
q
1
d
q
1
+
∂
h
3
ϕ
l
3
∂
q
2
d
q
2
+
∂
h
3
ϕ
l
3
∂
q
3
d
q
3
]
∧
d
q
3
{\displaystyle d\omega =[{\partial h_{1}\phi _{l_{1}} \over \partial q_{1}}dq_{1}+{\partial h_{1}\phi _{l_{1}} \over \partial q_{2}}dq_{2}+{\partial h_{1}\phi _{l_{1}} \over \partial q_{3}}dq_{3}]\land dq_{1}+[{\partial h_{2}\phi _{l_{2}} \over \partial q_{1}}dq_{1}+{\partial h_{2}\phi _{l_{2}} \over \partial q_{2}}dq_{2}+{\partial h_{2}\phi _{l_{2}} \over \partial q_{3}}dq_{3}]\land dq_{2}+[{\partial h_{3}\phi _{l_{3}} \over \partial q_{1}}dq_{1}+{\partial h_{3}\phi _{l_{3}} \over \partial q_{2}}dq_{2}+{\partial h_{3}\phi _{l_{3}} \over \partial q_{3}}dq_{3}]\land dq_{3}}
d
ω
=
[
∂
h
3
ϕ
l
3
∂
q
2
−
∂
h
2
ϕ
l
2
∂
q
3
]
d
q
2
d
q
3
+
[
∂
h
1
ϕ
l
1
∂
q
3
−
∂
h
3
ϕ
l
3
∂
q
1
]
d
q
3
d
q
1
+
[
∂
h
2
ϕ
l
2
∂
q
1
−
∂
h
1
ϕ
l
1
∂
q
2
]
d
q
1
d
q
2
{\displaystyle d\omega =[{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{2}}-{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{3}}]dq_{2}dq_{3}+[{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{3}}-{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{1}}]dq_{3}dq_{1}+[{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{1}}-{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{2}}]dq_{1}dq_{2}}
d
ω
=
1
h
2
h
3
[
∂
h
3
ϕ
l
3
∂
q
2
−
∂
h
2
ϕ
l
2
∂
q
3
]
dl
2
dl
3
+
1
h
1
h
3
[
∂
h
1
ϕ
l
1
∂
q
3
−
∂
h
3
ϕ
l
3
∂
q
1
]
dl
3
dl
1
+
1
h
1
h
2
[
∂
h
2
ϕ
l
2
∂
q
1
−
∂
h
1
ϕ
l
1
∂
q
2
]
dl
1
dl
2
{\displaystyle {\ce {d\omega ={1 \over h_{2}h_{3}}{[{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{2}}-{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{3}}]dl_{2}dl_{3}}+{1 \over h_{1}h_{3}}{[{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{3}}-{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{1}}]dl_{3}dl_{1}}+{1 \over h_{1}h_{2}}[{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{1}}-{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{2}}]dl_{1}dl_{2}}}}
d
ω
=
1
h
1
h
2
h
3
[
[
∂
h
3
ϕ
l
3
∂
q
2
−
∂
h
2
ϕ
l
2
∂
q
3
]
h
1
d
l
2
d
l
3
+
[
∂
h
1
ϕ
l
1
∂
q
3
−
∂
h
3
ϕ
l
3
∂
q
1
]
h
2
d
l
3
d
l
1
+
[
∂
h
2
ϕ
l
2
∂
q
1
−
∂
h
1
ϕ
l
1
∂
q
2
]
d
l
1
d
l
2
]
{\displaystyle d\omega ={1 \over h_{1}h_{2}h_{3}}[[{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{2}}-{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{3}}]h_{1}dl_{2}dl_{3}+[{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{3}}-{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{1}}]h_{2}dl_{3}dl_{1}+[{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{1}}-{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{2}}]dl_{1}dl_{2}]}
d
ω
=
1
h
1
h
2
h
3
|
h
1
q
1
^
h
2
q
2
^
h
3
q
3
^
∂
∂
q
1
∂
∂
q
2
∂
∂
q
3
h
1
ϕ
l
1
h
2
ϕ
l
2
h
3
ϕ
l
3
|
⋅
(
d
l
2
d
l
3
,
d
l
3
d
l
1
,
d
l
1
d
l
2
)
{\displaystyle d\omega ={1 \over h_{1}h_{2}h_{3}}{\begin{vmatrix}h_{1}{\hat {q_{1}}}&h_{2}{\hat {q_{2}}}&h_{3}{\hat {q_{3}}}\\{\partial \over \partial q_{1}}&{\partial \over \partial q_{2}}&{\partial \over \partial q_{3}}\\h_{1}\phi _{l_{1}}&h_{2}\phi _{l_{2}}&h_{3}\phi _{l_{3}}&\end{vmatrix}}\centerdot (dl_{2}dl_{3},dl_{3}dl_{1},dl_{1}dl_{2})}
d
ω
=
∇
x
ϕ
→
⋅
d
s
→
=
d
(
ϕ
→
⋅
d
l
→
)
{\displaystyle d\omega =\nabla {\text{x}}{\vec {\phi }}\centerdot d{\vec {s}}=d({\vec {\phi }}\centerdot d{\vec {l}})}
Generador de código CRC del modbus RTU a 16 bits [ editar ] Método 1 de la división con el polinomio generador [ editar ] Sea el dato : 0x1234
El polinomio generador del modbus es:
x
16
+
x
15
+
x
2
+
1
{\displaystyle x^{16}+x^{15}+x^{2}+1}
La condición inicial de el polinomio de modbus es 0xFFFF.
la división se realiza de la siguiente forma: se acomoda el dato comenzando con el bit menos significativo a la izquierda mientras que el polinomio generador con el bit mas significativo hacia la izquierda
Donde la primera fila corresponde a los datos, la segunda fila corresponde a la condición inicial la 4 fila es el resultado de la resta, la 5a fila es el polinomio generador CRC del modbus RTU
0
1
0
0
1
0
0
0
0
0
1
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
−
−
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−
−
−
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−
−
−
−
−
−
−
1
0
1
1
0
1
1
1
1
1
0
1
0
0
1
1
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
1
1
1
0
1
1
1
1
1
0
1
0
0
0
1
1
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
0
1
0
1
1
1
1
1
0
1
0
0
0
0
1
1
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
1
1
1
1
1
1
0
1
0
0
0
0
1
0
0
1
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
0
1
1
1
1
0
1
0
0
0
0
1
0
1
1
1
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
0
1
1
0
1
0
0
0
0
1
0
1
1
0
0
1
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
0
0
1
0
0
0
0
1
0
1
1
0
0
0
0
1
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
1
0
0
0
1
0
1
1
0
0
0
0
1
1
0
1
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
1
0
0
1
0
1
1
0
0
0
0
1
1
1
1
1
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
1
0
1
0
1
1
0
0
0
0
1
1
1
0
1
1
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
1
1
0
1
1
0
0
0
0
1
1
1
0
0
1
1
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
0
0
0
1
1
0
0
0
0
1
1
1
0
0
0
1
1
{\displaystyle {\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}0&1&0&0&1&0&0&0&0&0&1&0&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\hline 1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1\\-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\1&0&1&1&0&1&1&1&1&1&0&1&0&0&1&1&0\\1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\0&1&1&1&0&1&1&1&1&1&0&1&0&0&0&1&1&0\\&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&0&0&1&0&1&1&1&1&1&0&1&0&0&0&0&1&1&0&0\\&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&0&1&1&1&1&1&1&0&1&0&0&0&0&1&0&0&1&0\\&&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&&0&0&1&1&1&1&0&1&0&0&0&0&1&0&1&1&1&0&0\\&&&&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&&&&0&0&1&1&0&1&0&0&0&0&1&0&1&1&0&0&1&0&0\\&&&&&&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&&&&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&&&&&&0&0&0&1&0&0&0&0&1&0&1&1&0&0&0&0&1&0&0&0\\&&&&&&&&&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&&&&&&&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&&&&&&&&&0&1&0&0&0&1&0&1&1&0&0&0&0&1&1&0&1&0\\&&&&&&&&&&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&&&&&&&&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&&&&&&&&&&0&1&0&0&1&0&1&1&0&0&0&0&1&1&1&1&1&0\\&&&&&&&&&&&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&&&&&&&&&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&&&&&&&&&&&0&1&0&1&0&1&1&0&0&0&0&1&1&1&0&1&1&0\\&&&&&&&&&&&&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&&&&&&&&&&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&&&&&&&&&&&&0&1&1&0&1&1&0&0&0&0&1&1&1&0&0&1&1&0\\&&&&&&&&&&&&&&&1&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1\\&&&&&&&&&&&&&&&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-&-\\&&&&&&&&&&&&&&&0&0&0&1&1&0&0&0&0&1&1&1&0&0&0&1&1\\\end{array}}}
la ultima fila es el residuo que corresponde al valor empezando de derecha a izquierda CRC = 0xC70C.
Método 2 formato bit a bit. es un truco mas que un método. [ editar ] Procedimiento para generar el crc a 16 bits del modbus rtu
Cargue en un registro con el nombre de CRC con 0xFFF, correspondiente a la condición inicial
Realice X-Or con el primer byte del dato. y rellene con ceros la parte izquierda hasta completar los 16 bits del primer byte de datos antes de hacer la operación.
Realice un corrimiento del registro resultante hacia la derecha de un bit rellene con cero el bit entrante de la izquierda correspondiente al bit mas significativo y el bit que sale de la derecha correspondiente al bit menos significativo puede tener dos valores cero o uno.
Si es cero realice otro corrimiento.
Si es uno realize una X-Or con el polinomio generador acomodándolo de la siguiente forma 1010000000000001 donde el bit 17 no aparece en este polinomio. y se esta acomodando de tal forma que el bit menos significativo comienza en la izquierda y el bit mas significativo en la derecha desapareciendo el ultimo bit correspondiente a la posición 16.
Repita los pasos 3 y 4 hasta que se realicen 8 corrimientos.
cuando se haya terminado los 8 corrimientos se coloca el siguiente byte rellenando de ceros la parte izquierda del byte de datos hasta completar los 16 bits Ejemplo
Sea el dato : 0x1234
Descripción
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Condición inicial
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Primera palabra
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
0
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or
1
1
1
1
1
1
1
1
1
1
1
0
1
1
0
1
Corrimiento 1
0
1
1
1
1
1
1
1
1
1
1
1
0
1
1
0
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
1
0
1
1
1
1
1
1
1
1
1
0
1
1
1
Corrimiento 2
0
1
1
0
1
1
1
1
1
1
1
1
1
0
1
1
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
1
1
0
0
1
1
1
1
1
1
1
1
1
0
1
0
Corrimiento 3
0
1
1
0
0
1
1
1
1
1
1
1
1
1
0
1
0
Corrimiento 4
0
0
1
1
0
0
1
1
1
1
1
1
1
1
1
0
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
0
0
1
0
0
1
1
1
1
1
1
1
1
1
1
Corrimiento 5
0
1
0
0
1
0
0
1
1
1
1
1
1
1
1
1
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
1
1
0
1
0
0
1
1
1
1
1
1
1
1
0
Corrimiento 6
0
1
1
1
0
1
0
0
1
1
1
1
1
1
1
1
0
Corrimiento 7
0
0
1
1
1
0
1
0
0
1
1
1
1
1
1
1
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
0
0
1
1
0
1
0
0
1
1
1
1
1
1
0
Corrimiento 8
0
1
0
0
1
1
0
1
0
0
1
1
1
1
1
1
0 0x4D3F
Segunda palabra
0
0
0
0
0
0
0
0
0
0
1
1
0
1
0
0
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or
0
1
0
0
1
1
0
1
0
0
0
0
1
0
1
1
Corrimiento 1
0
0
1
0
0
1
1
0
1
0
0
0
0
1
0
1
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
0
0
0
0
1
1
0
1
0
0
0
0
1
0
0
Corrimiento 2
0
1
0
0
0
0
1
1
0
1
0
0
0
0
1
0
0
Corrimiento 3
0
0
1
0
0
0
0
1
1
0
1
0
0
0
0
1
0
Corrimiento 4
0
0
0
1
0
0
0
0
1
1
0
1
0
0
0
0
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
0
1
1
0
0
0
0
1
1
0
1
0
0
0
1
Corrimiento 5
0
1
0
1
1
0
0
0
0
1
1
0
1
0
0
0
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
1
1
1
1
0
0
0
0
1
1
0
1
0
0
1
Corrimiento 6
0
1
1
1
1
1
0
0
0
0
1
1
0
1
0
0
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
1
0
1
1
1
0
0
0
0
1
1
0
1
0
1
Corrimiento 7
0
1
1
0
1
1
1
0
0
0
0
1
1
0
1
0
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
1
0
0
1
1
1
0
0
0
0
1
1
0
1
1
Corrimiento 8
0
1
1
0
0
1
1
1
0
0
0
0
1
1
0
1
1
Polinomio
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
X-Or Polinomio
1
1
0
0
0
1
1
1
0
0
0
0
1
1
0
0
0xC70C
Estándar de programación [ editar ] https://bitbucket.org/Internetza/exampleplcequations/src/master/Estandar/Estandar.pdf
Sumatoria de números [ editar ]
(
x
−
1
)
k
=
∑
j
=
0
k
(
k
j
)
x
k
−
j
(
−
1
)
j
=
(
k
0
)
x
k
(
−
1
)
0
+
∑
j
=
1
k
(
k
j
)
x
k
−
j
(
−
1
)
j
=
x
k
+
∑
j
=
1
k
(
k
j
)
x
k
−
j
(
−
1
)
j
{\displaystyle (x-1)^{k}=\textstyle \sum _{j=0}^{k}\displaystyle {\binom {k}{j}}x^{k-j}(-1)^{j}={\binom {k}{0}}x^{k}(-1)^{0}+\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}x^{k-j}(-1)^{j}=x^{k}+\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}x^{k-j}(-1)^{j}}
Pasando ahora del lado izquierdo a la sumatoria
x
k
−
(
x
−
1
)
k
=
−
∑
j
=
1
k
(
k
j
)
x
k
−
j
(
−
1
)
j
{\displaystyle x^{k}-(x-1)^{k}=-\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}x^{k-j}(-1)^{j}}
si substituimos a x con los valores de 0, 1 , 2 ... , n
0
k
−
(
0
−
1
)
k
=
−
∑
j
=
1
k
(
k
j
)
0
k
−
j
(
−
1
)
j
{\displaystyle 0^{k}-(0-1)^{k}=-\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}0^{k-j}(-1)^{j}}
1
k
−
(
1
−
1
)
k
=
−
∑
j
=
1
k
(
k
j
)
1
k
−
j
(
−
1
)
j
{\displaystyle 1^{k}-(1-1)^{k}=-\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}1^{k-j}(-1)^{j}}
2
k
−
(
2
−
1
)
k
=
−
∑
j
=
1
k
(
k
j
)
2
k
−
j
(
−
1
)
j
{\displaystyle 2^{k}-(2-1)^{k}=-\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}2^{k-j}(-1)^{j}}
3
k
−
(
3
−
1
)
k
=
−
∑
j
=
1
k
(
k
j
)
3
k
−
j
(
−
1
)
j
{\displaystyle 3^{k}-(3-1)^{k}=-\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}3^{k-j}(-1)^{j}}
.
.
.
(
n
−
1
)
k
−
(
n
−
1
−
1
)
k
=
−
∑
j
=
1
k
(
k
j
)
(
n
−
1
)
k
−
j
(
−
1
)
j
{\displaystyle (n-1)^{k}-(n-1-1)^{k}=-\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}(n-1)^{k-j}(-1)^{j}}
n
k
−
(
n
−
1
)
k
=
−
∑
j
=
1
k
(
k
j
)
n
k
−
j
(
−
1
)
j
{\displaystyle n^{k}-(n-1)^{k}=-\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}n^{k-j}(-1)^{j}}
Sumando las filas
n
k
=
−
∑
j
=
1
k
(
k
j
)
(
−
1
)
j
S
k
−
j
{\displaystyle n^{k}=-\textstyle \sum _{j=1}^{k}\displaystyle {\binom {k}{j}}(-1)^{j}S_{k-j}}
Substituyendo k con 0, 1, 2, ..., hasta k
n
1
=
−
∑
j
=
1
1
(
1
j
)
(
−
1
)
j
S
1
−
j
=
S
0
{\displaystyle n^{1}=-\textstyle \sum _{j=1}^{1}\displaystyle {\binom {1}{j}}(-1)^{j}S_{1-j}=S_{0}}
n
2
=
−
∑
j
=
1
2
(
2
j
)
(
−
1
)
j
S
2
−
j
=
−
(
2
1
)
(
−
1
)
S
1
−
(
2
2
)
(
−
1
)
2
S
0
=
2
S
1
−
S
0
{\displaystyle n^{2}=-\textstyle \sum _{j=1}^{2}\displaystyle {\binom {2}{j}}(-1)^{j}S_{2-j}=-{\binom {2}{1}}(-1)S_{1}-{\binom {2}{2}}(-1)^{2}S_{0}=2S_{1}-S_{0}}
[
n
1
n
2
n
3
.
.
.
n
k
]
=
[
1
0
0
0
.
.
.
0
−
1
2
0
0
.
.
.
0
1
−
3
3
0
.
.
.
0
−
(
k
k
)
(
−
1
)
k
−
(
k
k
−
1
)
(
−
1
)
k
−
1
−
(
k
k
−
2
)
(
−
1
)
k
−
2
−
(
k
k
−
3
)
(
−
1
)
k
−
3
.
.
.
−
(
k
1
)
(
−
1
)
]
[
S
0
S
1
S
2
.
.
.
S
k
−
1
]
{\displaystyle {\begin{bmatrix}n^{1}\\n^{2}\\n^{3}\\...\\n^{k}\end{bmatrix}}={\begin{bmatrix}1&0&0&0&...&0\\-1&2&0&0&...&0\\1&-3&3&0&...&0\\-{\binom {k}{k}}(-1)^{k}&-{\binom {k}{k-1}}(-1)^{k-1}&-{\binom {k}{k-2}}(-1)^{k-2}&-{\binom {k}{k-3}}(-1)^{k-3}&...&-{\binom {k}{1}}(-1)\end{bmatrix}}{\begin{bmatrix}S_{0}\\S_{1}\\S_{2}\\...\\S_{k-1}\end{bmatrix}}}
[
S
0
S
1
S
2
.
.
.
S
k
−
1
]
=
[
1
0
0
0
.
.
.
0
−
1
2
0
0
.
.
.
0
1
−
3
3
0
.
.
.
0
−
(
k
k
)
(
−
1
)
k
−
(
k
k
−
1
)
(
−
1
)
k
−
1
−
(
k
k
−
2
)
(
−
1
)
k
−
2
−
(
k
k
−
3
)
(
−
1
)
k
−
3
.
.
.
−
(
k
1
)
(
−
1
)
]
−
1
[
n
1
n
2
n
3
.
.
.
n
k
]
{\displaystyle {\begin{bmatrix}S_{0}\\S_{1}\\S_{2}\\...\\S_{k-1}\end{bmatrix}}={\begin{bmatrix}1&0&0&0&...&0\\-1&2&0&0&...&0\\1&-3&3&0&...&0\\-{\binom {k}{k}}(-1)^{k}&-{\binom {k}{k-1}}(-1)^{k-1}&-{\binom {k}{k-2}}(-1)^{k-2}&-{\binom {k}{k-3}}(-1)^{k-3}&...&-{\binom {k}{1}}(-1)\end{bmatrix}}^{-1}{\begin{bmatrix}n^{1}\\n^{2}\\n^{3}\\...\\n^{k}\end{bmatrix}}}
Sea i la fila y j la columna entonces se puede reducir a la siguiente expresión
[
S
0
S
1
S
2
.
.
.
S
i
]
=
[
1
0
0
0
.
.
.
0
−
1
2
0
0
.
.
.
0
1
−
3
3
0
.
.
.
0
(
i
j
)
(
−
1
)
i
+
j
(
i
j
)
(
−
1
)
i
+
j
(
i
j
)
(
−
1
)
i
+
j
(
i
j
)
(
−
1
)
i
+
j
.
.
.
(
i
j
)
(
−
1
)
i
+
j
]
−
1
[
n
1
n
2
n
3
.
.
.
n
j
+
1
]
,
d
o
n
d
e
,
i
=
1
,
.
.
.
,
k
,
y
j
=
0
,
.
.
.
,
k
+
1
{\displaystyle {\begin{bmatrix}S_{0}\\S_{1}\\S_{2}\\...\\S_{i}\end{bmatrix}}={\begin{bmatrix}1&0&0&0&...&0\\-1&2&0&0&...&0\\1&-3&3&0&...&0\\{\binom {i}{j}}(-1)^{i+j}&{\binom {i}{j}}(-1)^{i+j}&{\binom {i}{j}}(-1)^{i+j}&{\binom {i}{j}}(-1)^{i+j}&...&{\binom {i}{j}}(-1)^{i+j}\end{bmatrix}}^{-1}{\begin{bmatrix}n^{1}\\n^{2}\\n^{3}\\...\\n^{j+1}\end{bmatrix}},donde,i=1,...,k,yj=0,...,k+1}
Entonces se puede reducir a
[
S
0
S
1
S
2
.
.
.
S
i
]
=
[
[
(
i
j
)
(
−
1
)
i
+
j
]
i
,
j
]
−
1
[
n
1
n
2
n
3
.
.
.
n
j
+
1
]
,
d
o
n
d
e
,
i
=
1
,
.
.
.
,
k
,
y
,
j
=
0
,
.
.
.
,
k
+
1
{\displaystyle {\begin{bmatrix}S_{0}\\S_{1}\\S_{2}\\...\\S_{i}\end{bmatrix}}={\begin{bmatrix}{\begin{bmatrix}{\binom {i}{j}}(-1)^{i+j}\end{bmatrix}}_{i,j}\end{bmatrix}}^{-1}{\begin{bmatrix}n^{1}\\n^{2}\\n^{3}\\...\\n^{j+1}\end{bmatrix}},donde,i=1,...,k,y,j=0,...,k+1}
Ejemplo :
[
S
0
S
1
]
=
[
1
0
−
1
2
]
−
1
[
n
1
n
2
]
=
[
2
/
2
0
1
/
2
1
/
2
]
[
n
1
n
2
]
=
[
n
n
(
n
+
1
)
/
2
]
{\displaystyle {\begin{bmatrix}S_{0}\\S_{1}\end{bmatrix}}={\begin{bmatrix}1&0\\-1&2\end{bmatrix}}^{-1}{\begin{bmatrix}n^{1}\\n^{2}\end{bmatrix}}={\begin{bmatrix}2/2&0\\1/2&1/2\end{bmatrix}}{\begin{bmatrix}n^{1}\\n^{2}\end{bmatrix}}={\begin{bmatrix}n\\n(n+1)/2\end{bmatrix}}}
Raíces de la función cuadrática [ editar ]
f
(
x
)
=
a
x
2
+
b
x
+
c
{\displaystyle f(x)=ax^{2}+bx+c}
f
(
x
′
+
x
0
)
=
a
(
x
′
+
x
0
)
2
+
b
(
x
′
+
x
0
)
+
c
=
a
x
′
2
+
(
2
a
x
0
+
b
)
x
′
+
a
x
0
2
+
b
x
0
+
c
=
a
x
2
+
f
(
x
0
)
{\displaystyle f(x\prime +x_{0})=a(x\prime +x_{0})^{2}+b(x\prime +x_{0})+c=ax\prime ^{2}+(2ax_{0}+b)x\prime +ax_{0}^{2}+bx_{0}+c=ax^{2}+f(x_{0})}
2
a
x
0
+
b
=
0
{\displaystyle 2ax_{0}+b=0}
x
0
=
−
b
/
2
a
{\displaystyle x_{0}=-b/2a}
f
(
x
′
+
x
0
)
=
a
x
′
2
+
f
(
x
0
)
=
0
{\displaystyle f(x\prime +x_{0})=ax\prime ^{2}+f(x_{0})=0}
x
′
2
=
−
f
(
x
0
)
/
a
{\displaystyle x\prime ^{2}=-f(x_{0})/a}
x
′
=
±
−
f
(
x
0
)
a
{\displaystyle x\prime =\pm {\sqrt {-f(x_{0}) \over a}}}
x
=
x
′
+
x
0
=
x
0
±
−
f
(
x
0
)
a
{\displaystyle x=x\prime +x_{0}=x_{0}\pm {\sqrt {-f(x_{0}) \over a}}}
Raíces de la función cúbica [ editar ]
f
(
x
)
=
a
x
3
+
b
x
2
+
c
x
+
d
=
0
{\displaystyle f(x)=ax^{3}+bx^{2}+cx+d=0}
f
(
x
′
+
x
0
)
=
a
(
x
′
+
x
0
)
3
+
b
(
x
′
+
x
0
)
2
+
c
(
x
′
+
x
0
)
+
d
=
0
{\displaystyle f(x\prime +x_{0})=a(x\prime +x_{0})^{3}+b(x\prime +x_{0})^{2}+c(x\prime +x_{0})+d=0}
f
(
x
′
+
x
0
)
=
a
(
x
′
3
+
3
x
0
2
x
′
+
3
x
0
x
′
2
+
x
0
3
)
+
b
(
x
′
2
+
2
x
0
x
′
+
x
0
2
)
+
c
(
x
′
+
x
0
)
+
d
=
0
{\displaystyle f(x\prime +x_{0})=a(x\prime ^{3}+3x_{0}^{2}x\prime +3x_{0}x\prime ^{2}+x_{0}^{3})+b(x\prime ^{2}+2x_{0}x\prime +x_{0}^{2})+c(x\prime +x_{0})+d=0}
f
(
x
′
+
x
0
)
=
a
x
′
3
+
(
3
a
x
0
+
b
)
x
′
2
+
(
3
a
x
0
2
+
2
b
x
0
+
c
)
x
′
+
a
x
0
3
+
b
x
0
2
+
c
x
0
+
d
=
0
{\displaystyle f(x\prime +x_{0})=ax\prime ^{3}+(3ax_{0}+b)x\prime ^{2}+(3ax_{0}^{2}+2bx_{0}+c)x\prime +ax_{0}^{3}+bx_{0}^{2}+cx_{0}+d=0}
f
(
x
′
+
x
0
)
=
a
x
′
3
+
f
″
(
x
0
)
x
′
2
+
f
′
(
x
0
)
x
+
f
(
x
0
)
=
0
{\displaystyle f(x\prime +x_{0})=ax\prime ^{3}+f''(x_{0})x\prime ^{2}+f'(x_{0})x+f(x_{0})=0}
3
a
x
0
+
b
=
0
{\displaystyle 3ax_{0}+b=0}
x
0
=
−
b
3
a
{\displaystyle x_{0}={\frac {-b}{3a}}}
f
(
x
′
+
x
0
)
=
a
x
′
3
+
f
′
(
x
0
)
x
+
f
(
x
0
)
=
0
{\displaystyle f(x\prime +x_{0})=ax\prime ^{3}+f'(x_{0})x+f(x_{0})=0}
(
u
+
v
)
3
=
u
3
+
3
u
2
v
+
3
u
v
2
+
v
3
=
u
3
+
3
u
v
(
u
+
v
)
+
v
3
{\displaystyle (u+v)^{3}=u^{3}+3u^{2}v+3uv^{2}+v^{3}=u^{3}+3uv(u+v)+v^{3}}
(
u
+
v
)
3
−
3
u
v
(
u
+
v
)
−
(
u
3
+
v
3
)
=
0
{\displaystyle (u+v)^{3}-3uv(u+v)-(u^{3}+v^{3})=0}
f
(
x
′
+
x
0
)
a
=
x
′
3
+
f
′
(
x
0
)
a
x
′
+
f
(
x
0
)
a
=
0
{\displaystyle {\frac {f(x'+x_{0})}{a}}=x'^{3}+{\frac {f'(x_{0})}{a}}x'+{\frac {f(x_{0})}{a}}=0}
x
′
=
u
+
v
{\displaystyle x'=u+v}
−
3
u
v
=
f
(
x
0
)
a
{\displaystyle -3uv={\frac {f(x_{0})}{a}}}
−
(
u
3
+
v
3
)
=
f
(
x
0
)
a
{\displaystyle -(u^{3}+v^{3})={\frac {f(x_{0})}{a}}}
Equivalencias
u
3
v
3
=
(
−
f
′
(
x
0
)
3
a
)
3
{\displaystyle u^{3}v^{3}=({\frac {-f'(x_{0})}{3a}})^{3}}
u
3
+
v
3
=
−
f
(
x
0
)
a
{\displaystyle u^{3}+v^{3}={\frac {-f(x_{0})}{a}}}
Manipulación
u
3
v
3
+
(
v
3
)
(
v
3
)
=
(
−
f
(
x
0
)
a
)
v
3
{\displaystyle u^{3}v^{3}+(v^{3})(v^{3})=({\frac {-f(x_{0})}{a}})v^{3}}
(
v
3
)
2
+
(
f
(
x
0
)
a
)
v
3
−
(
f
′
(
x
0
)
3
a
)
3
=
0
{\displaystyle (v^{3})^{2}+({\frac {f(x_{0})}{a}})v^{3}-({\frac {f'(x_{0})}{3a}})^{3}=0}
(
u
3
)
2
+
(
f
(
x
0
)
a
)
u
3
−
(
f
′
(
x
0
)
3
a
)
3
=
0
{\displaystyle (u^{3})^{2}+({\frac {f(x_{0})}{a}})u^{3}-({\frac {f'(x_{0})}{3a}})^{3}=0}
v
3
=
−
f
(
x
0
)
2
a
±
(
(
f
(
x
0
)
2
a
)
2
+
(
f
′
(
x
0
)
3
a
)
3
{\displaystyle v^{3}=-{\frac {f(x_{0})}{2a}}\pm {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}
u
3
=
−
f
(
x
0
)
2
a
∓
(
(
f
(
x
0
)
2
a
)
2
+
(
f
′
(
x
0
)
3
a
)
3
{\displaystyle u^{3}=-{\frac {f(x_{0})}{2a}}\mp {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}
x
′
=
u
+
v
=
−
f
(
x
0
)
2
a
∓
(
(
f
(
x
0
)
2
a
)
2
+
(
f
′
(
x
0
)
3
a
)
3
3
+
−
f
(
x
0
)
2
a
±
(
(
f
(
x
0
)
2
a
)
2
+
(
f
′
(
x
0
)
3
a
)
3
3
{\displaystyle x'=u+v={\sqrt[{3}]{-{\frac {f(x_{0})}{2a}}\mp {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}}+{\sqrt[{3}]{-{\frac {f(x_{0})}{2a}}\pm {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}}}
x
=
x
0
+
x
′
=
−
b
3
a
+
−
f
(
x
0
)
2
a
∓
(
(
f
(
x
0
)
2
a
)
2
+
(
f
′
(
x
0
)
3
a
)
3
3
+
−
f
(
x
0
)
2
a
±
(
(
f
(
x
0
)
2
a
)
2
+
(
f
′
(
x
0
)
3
a
)
3
3
{\displaystyle x=x_{0}+x'={\frac {-b}{3a}}+{\sqrt[{3}]{-{\frac {f(x_{0})}{2a}}\mp {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}}+{\sqrt[{3}]{-{\frac {f(x_{0})}{2a}}\pm {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}}}
Creación del Curso de Cálculo Integral en wikiversidad [ editar ] Curso de Cálculo Integral
Reconocimientos a mi labor [ editar ] Reconocimiento tal, 1234 . Como colaborador.[ 1] Referencias [ editar ]
↑ Si es público, en lo posible poner una referencia.
Esa ciudad, Tierra.![{\displaystyle dz={\partial z \over \partial (q_{1},q_{2},...,q_{n})}d(q_{1},q_{2},...,q_{n})=[H_{1}{\widehat {H}}_{1}H_{2}{\widehat {H}}_{2}...H_{n}{\widehat {H}}_{n}]dq}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09de85fc763d088ca556373166bf7492b56da270)
![{\displaystyle dz=[{\partial z \over H_{1}\partial q_{1}}{\partial z \over H_{2}\partial q_{2}}{...}{\partial z \over H_{n}\partial q_{n}}][{H_{1}dq_{1}}{H_{2}dq_{2}}{}...{H_{n}dq}]^{T}=[{\widehat {H_{1}}}{\widehat {H_{2}}}{...}{\widehat {H_{n}}}][{dl_{1}}{dl_{2}}{...}{dl_{n}}]^{T}=\nabla _{l}zdl}](https://wikimedia.org/api/rest_v1/media/math/render/svg/418779914d7502985fd18bf22e7aed4a078d63d9)
![{\displaystyle d\omega =[{\partial h_{1}\phi _{l_{1}} \over \partial q_{1}}dq_{1}+{\partial h_{1}\phi _{l_{1}} \over \partial q_{2}}dq_{2}+{\partial h_{1}\phi _{l_{1}} \over \partial q_{3}}dq_{3}]\land dq_{1}+[{\partial h_{2}\phi _{l_{2}} \over \partial q_{1}}dq_{1}+{\partial h_{2}\phi _{l_{2}} \over \partial q_{2}}dq_{2}+{\partial h_{2}\phi _{l_{2}} \over \partial q_{3}}dq_{3}]\land dq_{2}+[{\partial h_{3}\phi _{l_{3}} \over \partial q_{1}}dq_{1}+{\partial h_{3}\phi _{l_{3}} \over \partial q_{2}}dq_{2}+{\partial h_{3}\phi _{l_{3}} \over \partial q_{3}}dq_{3}]\land dq_{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5f4f3d36c61d2060b0bee7f1f95685548cb7d55)
![{\displaystyle d\omega =[{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{2}}-{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{3}}]dq_{2}dq_{3}+[{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{3}}-{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{1}}]dq_{3}dq_{1}+[{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{1}}-{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{2}}]dq_{1}dq_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/522fdb6dc4d5ca907bad79afb788c6190e348f09)
![{\displaystyle {\ce {d\omega ={1 \over h_{2}h_{3}}{[{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{2}}-{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{3}}]dl_{2}dl_{3}}+{1 \over h_{1}h_{3}}{[{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{3}}-{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{1}}]dl_{3}dl_{1}}+{1 \over h_{1}h_{2}}[{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{1}}-{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{2}}]dl_{1}dl_{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2aa614cf2527408c87336eb9c18b1eb32077204e)
![{\displaystyle d\omega ={1 \over h_{1}h_{2}h_{3}}[[{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{2}}-{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{3}}]h_{1}dl_{2}dl_{3}+[{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{3}}-{{\partial h_{3}\phi _{l_{3}}} \over \partial q_{1}}]h_{2}dl_{3}dl_{1}+[{{\partial h_{2}\phi _{l_{2}}} \over \partial q_{1}}-{{\partial h_{1}\phi _{l_{1}}} \over \partial q_{2}}]dl_{1}dl_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cfb31a0bc80f7494f4673081eb9e4e9c961532b)
![{\displaystyle x'=u+v={\sqrt[{3}]{-{\frac {f(x_{0})}{2a}}\mp {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}}+{\sqrt[{3}]{-{\frac {f(x_{0})}{2a}}\pm {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df236fbd57ba13767ab3fc359ad5765097d2ff14)
![{\displaystyle x=x_{0}+x'={\frac {-b}{3a}}+{\sqrt[{3}]{-{\frac {f(x_{0})}{2a}}\mp {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}}+{\sqrt[{3}]{-{\frac {f(x_{0})}{2a}}\pm {\sqrt {({\frac {(f(x_{0})}{2a}})^{2}+({\frac {f'(x_{0})}{3a}})^{3}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3d89134a8264453882d266458af956d4f572f72)
